Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{5}{\sqrt{9-4 x^{2}}} d x $$

Step-by-Step Solution

Answer

\( \frac{5}{2} \sin^{-1} \left( \frac{2x}{3} \right) + C \)

Step 1: Recognize the Integral Form

The integral \( \int \frac{5}{\sqrt{9-4 x^{2}}} \, dx \) matches a standard integral form involving \( \sin^{-1}(x) \). The form \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C \) can be used, where \( a^2 = 9 \) and hence \( a = 3 \).

Step 2: Rewrite the Integral

First, adjust the integral to match the standard form more closely. Rewrite the integral as \( \int \frac{5}{\sqrt{9 - (2x)^2}} \, dx \). Recognize that the expression under the square root can be viewed as \( a^2 - (bx)^2 \) where \( a = 3 \) and \( b = 2 \).

Step 3: Use the Substitution Method

Let \( u = 2x \). Then, \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \). Substitute \( u = 2x \) and \( dx \) to transform the integral into \( \int \frac{5}{\sqrt{9 - u^2}} \cdot \frac{1}{2} \, du = \frac{5}{2} \int \frac{1}{\sqrt{9-u^2}} \, du \).

Step 4: Apply the Arcsine Integral Formula

Now the integral is in the form \( \int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1} \left( \frac{u}{a} \right) + C \). For our integral, \( a = 3 \), so \( \int \frac{1}{\sqrt{9 - u^2}} \, du = \sin^{-1}\left( \frac{u}{3} \right) + C \).

Step 5: Back Substitute for Original Variable

Replace \( u = 2x \) back into the solution to return to the variable in the original problem. Therefore, the integral becomes \( \frac{5}{2} \left[ \sin^{-1} \left( \frac{2x}{3} \right) \right] + C \).

Step 6: Write the Final Answer

The final evaluated integral is \[ \int \frac{5}{\sqrt{9-4x^2}} \, dx = \frac{5}{2} \sin^{-1} \left( \frac{2x}{3} \right) + C \].

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's a toolset for analyzing change and motion. Calculus emerges mainly in two parts: Differential Calculus, concerned with the rate of change — essentially the slope at a point — and Integral Calculus, which deals with the accumulation of quantities, like areas under curves.
The process of integration helps find the total amount or accumulated change. You can think of integrals as the reverse operation of a derivative. In some problems, especially those involving non-straightforward functions, particular techniques are used, such as substitution, partial fractions, or integration by parts.
Key to learning calculus is understanding how these different pieces fit together. It's about building a toolkit for breaking down complex mathematical problems into solvable parts. The beauty lies in recognizing patterns and applying standard forms, like transforming and reducing problems using relevant theorems and formulas.

Trigonometric Substitution

Trigonometric substitution is a specific method in integration. It leverages trigonometric identities to simplify integrals involving square roots. Typically, these integrals have the form \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \).
The goal is to replace the variable with a trigonometric function that simplifies the expression into a more manageable form. For example:

• For \( \sqrt{a^2 - x^2} \), use \( x = a \sin(\theta) \)
• For \( \sqrt{a^2 + x^2} \), use \( x = a \tan(\theta) \)
• For \( \sqrt{x^2 - a^2} \), use \( x = a \sec(\theta) \)

These substitutions convert a challenging integral into one involving simpler trigonometric functions. The accompanying derivatives help transform the entire expression. After substitution, the trigonometric integral can typically be solved using known formulas or simpler calculus techniques. The solution is then reverted back to the original variable's terms. Remember, trigonometric substitution is most useful when dealing with integrals featuring these root expressions.

Inverse Trigonometric Functions

Inverse trigonometric functions are the 'undo' of regular trigonometric functions. They find the angle with a given trigonometric ratio. Common inverse functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).
In integration, these functions often appear when dealing with expressions entailing the form \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \). Recalling from trigonometry:

• \( \sin^{-1}(x/a) \) gives the angle whose sine is \( x/a \)
• \( \cos^{-1}(x/a) \) does the same for the cosine
• \( \tan^{-1}(x/a) \) for the tangent

To find these antiderivatives, integral solutions use the derived formula connecting these inverse functions with their patterns. For instance, \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \) directly uses \( \sin^{-1}(x/a) \). Knowing these forms aids in identifying and simplifying complex calculus problems, making inverse trigonometric functions vital for tackling various challenging integrals efficiently.